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Statistics, 2ADV S3 2025 MET1 8

Consider

\begin{align*}
f(x)=\left\{\begin{array}{cc}
\dfrac{3}{8}(4-3 x) & 0 \leq x \leq \dfrac{4}{3} \\
0 & \text {otherwise }
\end{array}\right.
\end{align*}

  1. The continuous random variable \(X\) has probability density function \(f(x)\).
  2. Find \(k\) such that  \(\operatorname{Pr}(X>k)=\dfrac{9}{16}\).   (3 marks)

  3. The function \(h(x)\) is a transformation of \(f(x)\) such that
  4. \begin{align*}
    \ \ \ \ h(x)=m f(x)+n
    \end{align*}
  5. where \(m\) and \(n\) are real numbers.
  6. Find  \(\displaystyle \int_0^{\tfrac{4}{3}} h(x) d x\)  in terms of \(m\) and \(n\).    (2 marks)

Show Answers Only

a.    \(k=\dfrac{1}{3}\)

b.    \(\displaystyle\int_0^{\frac{4}{3}} h(x)=m+\dfrac{4}{3} n\)

Show Worked Solution

a.    \(\text{Solve}\ \ \operatorname{Pr}(X>k)=\dfrac{9}{16}\ \ \text{for}\ k\) :

\(\operatorname{Pr}(X>k)\) \(=\displaystyle\int_k^{\frac{4}{3}} \frac{3}{8}(4-3 x) d x\)
  \(=\displaystyle-\frac{3}{8} \int_k^{\frac{4}{3}}(3 x-4) d x\)
  \(=-\dfrac{3}{8} \cdot \dfrac{1}{2} \cdot \dfrac{1}{3}\left[(3 x-4)^2\right]_k^{\frac{4}{3}}\)
  \(=-\dfrac{1}{16}\left[(4-4)^2-(3 k-4)^2\right]\)
  \(=\dfrac{(3 k-4)^2}{16}\)
♦ Mean mark (a) 47%.

 

\(\dfrac{(3 k-4)^2}{16}\) \(=\dfrac{9}{16}\)
\((3 k-4)^2\) \(=9\)
\(3 k-4\) \(= \pm 3\)

 

\(3 k=1 \ \ \ \ \text{or}\ \ \ \ 3 k=7\)

\(k=\dfrac{1}{3} \quad \quad \ \ \ k=\dfrac{7}{3}\ \ \left(\text{No solution as}\ k \in\left[0, \dfrac{4}{3}\right]\right)\)

 \(\therefore\ k=\dfrac{1}{3}\)
  

b.    \(h(x)=m f(x)+n\)

  \(\displaystyle\int_0^{\frac{4}{3}} h(x)\) \(=\displaystyle \int_0^{\frac{4}{3}}(m f(x)+n) d x\)
    \(=m \underbrace{\displaystyle \int_0^{\frac{4}{3}} f(x) d x}_{=1\  \ \text{since p.d.f. }}+n \displaystyle\int_0^{\frac{4}{3}} 1\ d x\)
    \(=m+n[x]_0^{\frac{4}{3}}\)
    \(=m+\dfrac{4}{3} n\)
♦♦ Mean mark (b) 38%.